Is $A$ nilpotent matrix? Let $A$ be a $n \times n$ matrix over $\mathbb{C}$ and $I+At$ is invertible for all t, does that imply $A$ is nilpotent?
I know the converse is true, but is it also true?
 A: Mariano's answer is definitely the "right" way of doing it, but I thought I'd add a geometric answer, since you're working over $\mathbb{C}$.
By hypothesis, you have a global holomorphic map $\mathbb{C} \rightarrow GL_n(\mathbb{C})$ sending $t$ to $I+tA$ (with no assumption on $A$, the codomain would only be $M_n(\mathbb{C})$ or instead you could restrict the domain to a small neighborhood of 0 and leave the codomain $GL_n$). You can compose this with the inverting map $GL_n{\mathbb{C}} \rightarrow GL_n{\mathbb{C}}$ followed by the obvious inclusion $GL_n{(\mathbb{C})} \rightarrow \mathbb{C}^{n^2}$ to get a global holomorphic function $S:\mathbb{C} \rightarrow \mathbb{C}^{n^2}$. Locally near $t=0$, it is clear that $S$ is given by the power series $S'=I-At+A^2t^2-A^3t^3+...$. A property of holomorphic functions on disks is that any local power series expansion around the center must hold on the whole disk, so $S'=S$ on all of $\mathbb{C}$. In particular, $S'$ is absolutely convergent for all $t \in \mathbb{C}$. But taking $t$ with $||t||>>0$, it is clear (intuitively, but see comments) that this forces $A^N=0$ for sufficiently large $N$. 
A: If $I+At$ is invertible for all $t$, then $t^{-1}I+A=t^{-1}(I+At)$ is invertible for all $t\neq0$, so that $\det(sI+A)\neq0$ for all nonzero $s$, so that the only root of the characteristic polynomial $p$ of $A$ is zero, so that $p(\lambda)=\lambda^n$, so that $A$ is nilpotent, according to the Cayley-Hamilton theorem.
